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Essential Cheat Sheet: Financial Contents
WEEK 1 - Yield curve and fixed income instruments ................................................ 2
Markets and Institutions WEEK 2 - Monetary policy, inflation, yield curve inversion ...................................... 5
WEEK 3 - FX markets: instruments, theoretical parities, and their violation ........... 7
WEEK 4 – Banking, regulation and market liquidity ................................................. 9
Author: Linh Nguyen WEEK 5 – Asset management ................................................................................ 10
Study year: 2023 – 2024
Course name: Financial Markets and Institutions
Vrije Universiteit Amsterdam
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WEEK 1 - Yield curve and fixed income instruments Continuous compounding:
Yield curve: Collection of Rf interest rates for different maturities {𝑟(0, 𝑇)} 𝑇 𝑒 −(𝑇2−𝑇1)×𝑓(0,𝑇1,𝑇2) = 𝐹(0, 𝑇1 , 𝑇2 ) = e−𝑟(0,𝑇2)×𝑇2+𝑟(0,𝑇1)×𝑇1
• Upward sloping: Normal curve. S/T IR < L/T IR. Investors demand higher return for holding L/T − ln 𝐹(0, 𝑇1 , 𝑇2 )
securities. 𝑓(0, 𝑇1 , 𝑇2 ) =
𝑇2 − 𝑇1
• Flat: Investors are indifferent.
• Inverted: S/T IR > L/T IR. Signs of recession or that S/T rates are expected to go lower. 2. Basics of interest-rate risk management
Yield to maturity (YTM): the annual expected return of a bond if held until maturity, also referred to P: price of financial instruments; r: interest rates
as book yield. If IRs change by Δ𝑟, then price changes by Δ𝑃. The relationship (absolute) is
𝑁 1 ′′
𝑦 −𝑛𝑇𝑖 Δ𝑃 ≈ 𝑃𝑟′ ∗ Δ𝑟 + 𝑃𝑟𝑟 ∗ (Δ𝑟)2
𝑃(𝑡, 𝑇) = ∑ 𝐶𝐹𝑖 (1 + ) 2
𝑛 ′ ′′
𝑖=1 𝑃𝑟 and 𝑃𝑟 are the first and second derivative of P w.r.t. r.
Bootstrapping the curve: Constructing a (zero coupon) yield curve from coupon bearing products The relationship (relative, %) is
like coupon bonds or swaps, FRAs. Δ𝑃 1 1 1 ′′
Example: IR r1(0,1) associated with discount factor ≈ − (− 𝑃𝑟′ ) Δ𝑟 + ( 𝑃𝑟𝑟 ) (Δ𝑟)2
𝑃 𝑃 2 𝑃
1 1 Δ𝑃 1
𝑍(0,1) = ⇒ 𝑟1 (0,1) = − 1 ≈ 0.0417 = 4.17% = 417𝑏. 𝑝. = −𝐷Δ𝑟 + 𝐶(Δ𝑟)2
1 + 𝑟1 (0,1) 𝑍(0,1) 𝑃 2
1
Duration: 𝐷 = − 𝑃𝑟′ (minus 1 over P times first derivative of P w.r.t. r)
𝑃
N-times compounded IR, 𝒓𝒏 (𝟎, 𝑻𝒊 ) Continuous compounding IR 𝒓(𝟎, 𝑻𝒊 ): 𝒏 = ∞ 1′′ (1 over P times second derivative of P w.r.t. r)
Convexity: 𝐶 = 𝑃𝑟𝑟
−𝑛𝑇𝑖 𝑃
𝑟𝑛 (0, 𝑇𝑖 )
𝑍(0, 𝑇𝑖 ) = (1 + ) 𝑍(0, 𝑇𝑖 ) = 𝑒 −𝑟(0,𝑇𝑖)𝑇𝑖
𝑛 DURATION AND CONVEXITY OF ZERO-COUPON BOND
1 ln 𝑍(0, 𝑇)
𝑟𝑛 (0, 𝑇) = 𝑛 (𝑍(0, 𝑇)−𝑛𝑇 − 1) 𝑟(0, 𝑇) = − Time-t price of zero-coupon bond, maturity T is
𝑇
𝒁(𝒕, 𝑻) = 𝑒 −𝑟(𝑡,𝑇)(𝑇−𝑡)
𝑟(𝑡, 𝑇) is continuously compounded interest rate.
Discount curve: A collection of zero-coupon bond prices {𝑍(0, 𝑇)] 𝑇 for different maturities
Duration of zero-coupon bond = time to maturity D = T – t
Duration measures the sensitivity of prices to parallel shifts of the yield curve.
Forward rate agreement (FRA): noncash contract between two counterparties
Convexity of zero-coupon bond = time to maturity squared. C = (T – t )2
• Time 0, Notional amount N and the forward rate 𝑓𝑛 (0, 𝑇1 , 𝑇2 ) are agreed
• Time 𝑇1 < 𝑇2 , 𝑟𝑛 (𝑇1 , 𝑇2 ) is revealed.
DURATION AND CONVEXITY OF COUPON BOND (CP BOND)
• Time T2:
CP bond price P(t,T), CP payment times Ti : i = 1, 2, …, n; Tn = T and payments CFi
o Buyer needs to pay 𝑁 ∗ Δ𝑓𝑛 (0, 𝑇1 , 𝑇2 )
Or Time-t price of CP bond, maturity T
o Seller needs to give 𝑁 ∗ Δ𝑟𝑛 (𝑇1 , 𝑇2 ) 𝑛 𝑛
1 𝑃(𝑡, 𝑇𝑖 ) = ∑ 𝐶𝐹𝑖 ∗ 𝑍(𝑡, 𝑇𝑖 ) = ∑ 𝐶𝐹𝑖 𝑒 −𝑟(𝑡,𝑇𝑖 )(𝑇𝑖 −𝑡)
Δ≡
𝑛 𝑖=1 𝑖=1
• So, the parties exchange the dollar difference between two rates at maturity T2: 𝑛
𝑦 −𝑛𝑇𝑖
𝑁 ∗ Δ[𝑟𝑛 (𝑇1 , 𝑇2 ) − 𝑓𝑛 (0, 𝑇1 , 𝑇2 )] 𝑃(𝑡, 𝑇𝑖 ) = ∑ 𝐶𝐹𝑖 ∗ (1 + )
𝑛
To ensure that the FRA costs nothing at inception, forward rate 𝑓𝑛 (0, 𝑇1 , 𝑇2 ) must be set as: 𝑖=1
where n = number of cashflow per year; y = YTM
𝑛(𝑇2 −𝑇1 ) 𝑍(0, 𝑇1 )
𝑓𝑛 (0, 𝑇1 , 𝑇2 ) = [−1 + √ ]∗𝑛 Equals to sum of all CF at time i times time-t price of ZC bond, maturity time i. Z here is also the
𝑍(0, 𝑇2 ) discount factor at time t.
𝑓𝑛 (0,0, 𝑇) ≡ 𝑟𝑛 (0, 𝑇) If there is a parallel shift in the continuously compounded yield curve, such that regardless of
Forward discount factor: maturity, rates change by the same amount: Δ𝑟(𝑡, 𝑇𝑖 ) = Δ𝑟 ∀𝑖 then
−𝑛(𝑇2 −𝑇1 ) 𝑛 𝑛
𝑓𝑛 (0, 𝑇1 , 𝑇2 ) 𝑍(0, 𝑇2 ) 𝐶𝐹𝑖 𝑍(𝑡, 𝑇𝑖 )
𝐹(0, 𝑇1 , 𝑇2 ) = (1 + ) = 𝐷 = ∑ 𝑤𝑖 (𝑇𝑖 − 𝑡) = ∑ (𝑇𝑖 − 𝑡)
𝑛 𝑍(0, 𝑇1 ) 𝑃
𝑖=1 𝑖=1
